\newproblem{lay:1_2_33}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.2.33}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Find the interpolating polynomial $p(t)=a_0+a_1t+a_2t^2$ for the data $(1,6)$, $(2,15)$, and $(3,28)$.
}
{
  % Solution
	We need to find $a_0$, $a_1$ and $a_2$ such that
	\begin{center}
		$\begin{array}{rcl}
		   a_0+a_1(1)+a_2(1)^2&=&6\\
		   a_0+a_1(2)+a_2(2)^2&=&15\\
		   a_0+a_1(3)+a_2(3)^2&=&28\\
		\end{array}
		\Rightarrow
		\left(\begin{array}{ccc}1 & 1 & 1^2 \\ 1 & 2 & 2^2 \\ 1 & 3 & 3^2\end{array}\right)
		\left(\begin{array}{c}a_0\\a_1\\a_2\end{array}\right)=
		\left(\begin{array}{c}6\\15\\28\end{array}\right)
		$
	\end{center}
	The augmented matrix is
	\begin{center}
		$\left(\begin{array}{ccc|c}1 & 1 & 1^2 & 6\\ 1 & 2 & 2^2 & 15\\ 1 & 3 & 3^2 & 28\end{array}\right) \sim
		 \left(\begin{array}{ccc|c}1 & 0 & 0 & 1\\ 0 & 1 & 0 & 3\\ 0 & 0 & 1 & 2\end{array}\right)$
	\end{center}
	Consequently, $a_0=1$, $a_1=3$ and $a_2=2$. The interpolating polynomial is
	\begin{center}
		$p(t)=1+3t+2t^2$
	\end{center}
	The data points as well as the polynomial are represented below
	\begin{center}
		\includegraphics[scale=0.6]{Tema2/lay_1_2_33.eps}
	\end{center}
}
\useproblem{lay:1_2_33}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
